Work Required to Empty Tank of Hot Chocolate

AI Thread Summary
The discussion revolves around calculating the work required to empty a tank shaped like an inverted right circular cone filled with hot chocolate. The tank has a height of 8 meters and a radius of 6 meters, with 7 meters of hot chocolate inside. The initial approach to the problem involved incorrect integration, leading to confusion about the correct formula. The correct method involves integrating the force, which is the mass of the hot chocolate multiplied by gravity, over the distance it needs to be pumped out. The final calculation yields a work value of approximately 8.229 million joules.
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A tank in the shape of an inverted right circular cone has height 8 meters and radius 6 meters. It is filled with 7 meters of hot chocolate.
Find the work required to empty the tank by pumping the hot chocolate over the top of the tank. Note: the density of hot chocolate is 1510kg/m^3

m = pi*r^2
m = 36*pi*1510
m = 54360*pi
distance = (8-x)

\int_{0}^{7} 54360*pi*(8-x) dx

this is clearly wrong, because i already tried to submit this, but it wouldn't accept it. can someone tell me what I am doing wrong?
 
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ProBasket said:
A tank in the shape of an inverted right circular cone has height 8 meters and radius 6 meters. It is filled with 7 meters of hot chocolate.
Find the work required to empty the tank by pumping the hot chocolate over the top of the tank. Note: the density of hot chocolate is 1510kg/m^3
From problem statement:
{Height of Liquid Top Surface} = h
{Radius of Liquid Top Surface} = r = (6/8)*h = (3/4)*h
{Volume of Liquid in Tank} = (1/3)*π*r2*h = (1/3)*π{(3/4)*h}2*h = (0.589)*h3
{Mass of Liquid in Tank} = M = {Density}*{Volume of Liquid in Tank} =
= (1510)*(0.589)*h3 = (889.4)*h3

{dM/dh} = (3)*(889.4)*h2 = (2668)*h2
{Work Lifting Liquid Over Tank Top} = W = ∫ g*(8 - h)*(dM/dh)*dh

\ \ \ \ W = \int_{0}^{7} g(8 - h)(2668)h^{2} \ dh

\ \ \ \ W = (2668)g\int_{0}^{7} (8 - h)h^{2} \ dh

\ \ \ \ W = (26173)\int_{0}^{7} (8h^{2} - h^{3}) \ dh

\ \ \ \ \color{red} W = (26173)(314.4) = (8229000 \ J) = (8.229*10^{6} \ J)


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Your approach is correct, but there is a mistake in your integration. The correct integral should be:

\int_{0}^{7} 54360*pi*(8-x)*1510 dx

This is because the work required to pump out the hot chocolate is equal to the force (mass * gravity) multiplied by the distance it is pumped (8-x) and the density of the hot chocolate. So the formula should be:

Work = Force * Distance = (m*g) * (8-x) = (m*1510) * (8-x) = (36*pi*1510) * (8-x)

Then you integrate this formula from 0 to 7 (the initial and final heights of the hot chocolate) and you will get the correct answer.
 
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